For an objective function Z=ax+by , where a, b > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:

Options

A. b-3a=0
B. a=3b
C. a+2b=0
D. 2a-b=0

Answer

To solve the problem, we need to determine the condition on \( a \) and \( b \) such that the maximum value of the objective function \( Z = ax + by \) occurs at both points \( (30, 30) \) and \( (0, 40) \).

### Step-by-Step Solution:

1. **Evaluate the Objective Function at the Points:**
- For point \( (30, 30) \):
\[ Z_1 = a(30) + b(30) = 30a + 30b \]
- For point \( (0, 40) \):
\[ Z_2 = a(0) + b(40) = 40b \]

2. **Set the Objective Function Values Equal:**
Since we want the maximum \( Z \) to occur at both points, we set \( Z_1 \) equal to \( Z_2 \):
\[ 30a + 30b = 40b \]

3. **Rearrange the Equation:**
Rearranging the equation gives:
\[ 30a + 30b - 40b = 0 \]
\[ 30a - 10b = 0 \]

4. **Solve for \( a \) in terms of \( b \):**
Dividing the entire equation by 10:
\[ 3a - b = 0 \]
Thus, we can express \( b \) in terms of \( a \):
\[ b = 3a \]

5. **Conclusion:**
The condition on \( a \) and \( b \) such that the maximum \( Z \) occurs at both points \( (30, 30) \) and \( (0, 40) \) is:
\[ b = 3a \quad \text{(where \( a, b > 0 \))} \]